Báo cáo toán học: "An Analogue of Covering Space Theory for Ranked Posets" pot

We callP a weighted-relation poset if all the covering relations ofP are assigned a positive integer weight.. If P is a poset of “natural” combinatorial objects, the elements of its univ

An Analogue of Covering Space Theory for Ranked

Posets Michael E Hoffman

Dept of Mathematics

U S Naval Academy, Annapolis, MD 21402

meh@usna.edu Submitted: May 10, 2001 Accepted: October 11, 2001

MR Classifications: Primary 06A07,05A15; Secondary 57M10

Abstract

Suppose P is a partially ordered set that is locally finite, has a least element,

and admits a rank function We callP a weighted-relation poset if all the covering

relations ofP are assigned a positive integer weight We develop a theory of covering

maps for weighted-relation posets, and in particular show that any weighted-relation posetP has a universal cover e P → P , unique up to isomorphism, so that

1 eP → P factors through any other covering map P 0 → P ;

2 every principal order ideal of eP is a chain; and

3 the weight assigned to each covering relation of eP is 1.

If P is a poset of “natural” combinatorial objects, the elements of its universal

cover eP often have a simple description as well For example, if P is the poset of

partitions ordered by inclusion of their Young diagrams, then the universal cover eP

is the poset of standard Young tableaux; if P is the poset of rooted trees ordered

by inclusion, then eP consists of permutations We discuss several other examples,

including the posets of necklaces, bracket arrangements, and compositions

For topological spaces, the notion of a covering space is familiar (see, e.g., [9]): a covering

map p : X 0 → X is a continuous surjection such that, for sufficiently small open sets U ⊂

X, p −1 (U) is a disjoint union of open sets in X 0 each of which p maps homeomorphically onto U For any space X satisfying appropriate hypotheses (e.g., that X is connected,

locally arcwise connected and semilocally simply connected), there is a simply connected

covering space π : e X → X, which is universal in the sense that it “factors through” any

other connected cover of X, i.e., if p : X 0 → X is any covering map with X 0 conneced,

then there is a covering map f : e X → X 0 so that π = pf The universal covering space of

X is unique up to homeomorphism over X.

In this paper we develop a theory of covering maps for ranked posets More precisely,

we define covering maps of “weighted-relation” posets, which are locally finite ranked posets with least element that have a positive integer weight associated with each of their

covering relations We show that every such weighted-relation poset P has a universal

cover eP → P , unique up to isomorphism in an appropriate category, which factors through

any other cover P 0 → P The universal cover e P is “simple” in the sense that its Hasse

diagram is a tree and all its covering relations have weight 1

In many cases where P is a poset of familiar combinatorial objects, the elements of the

universal cover eP also have a simple description For example, the poset of monomials

in commuting variables x1, , x k has a universal cover whose elements are monomials in

k noncommuting variables (Example 2 in §4 below); the poset of compositions (with an

appropriate choice of weights) has as its universal cover the poset of Cayley permutations

in the sense of [6] (Example 6) We discuss several other examples, including the posets

of necklaces, bracket arrangements, partitions, and rooted trees

Our terminology for posets follows [11] Let (P, ) be a locally finite poset with least

element ˆ0 and rank function | · | By a weight system on the relations of P , we mean a

function n that assigns a nonnegative integer n(x, y) to every pair x, y ∈ P so that

1 n(x, y) 6= 0 if and only if x  y;

2 for all elements x ≺ y and nonnegative integers |x| ≤ k ≤ |y|,

n(x, y) = X

|z|=k

n(x, z)n(z, y).

(Note that the second condition implies n(x, x) = 1 for all x ∈ P )

We call a poset P together with a weight system on its relations a weighted-relation

poset By induction on |y| − |x| it is easy to prove from the definition that for any x ≺ y

in P

n(x, y) = X

x=x1≺x2≺···≺x k =y

n(x1, x2)n(x2, x3)· · · n(x k−1 , x k ),

where the sum is over all saturated chains x = x1 ≺ x2 ≺ · · · ≺ x k = y from x to y: thus,

to define n it suffices to give n(x, y) when y covers x In particular, any ranked, locally

finite poset with least element can be made a weighted-relation poset by assigning 1 to every covering relation

The motivation for this definition comes from thinking of a covering relation x ≺ y of

P as indicating y can be built from x by some kind of elementary operation: n(x, y) is

the number of ways this can be done Then in general n(u, v) is the number of ways that

v can be built up from u via a sequence of elementary operations For examples see §4

below

Let W be the category whose objects are weighted-relation posets, and whose

mor-phisms are defined as follows A morphism of weighted-relation posets P , P 0 is a

rank-preserving function f : P → P 0 such that, for any elements t, s of P ,

n(f (t), f (s)) ≥ X

s 0 ∈f −1 (f(s))

n(t, s 0 ). (1)

In particular, any such function f is order-preserving Also, if f has an inverse f −1that is

also a morphism of weighted-relation posets, then n(f (t), f (s)) = n(t, s) for all t, s ∈ P

We call a weighted-relation poset P simple if n(x, y) is 1 or 0 for any x, y ∈ P The

following result is evident

Proposition 2.1 If P is a weighted-relation poset, the following are equivalent:

(i) P is simple;

(ii) the Hasse diagram of P is a tree, and every covering relation has weight 1;

(iii) for every x ∈ P , n(ˆ0, x) = 1.

We also record the following fact, which is an immediate consequence of inequality (1)

simple, then f is an injective function and P is simple.

Let P 0 and P be weighted-relation posets We say that a rank-preserving function π :

P 0 → P is a covering map if, whenever s, r ∈ P with π(s 0 ) = s,

n(s, r) = X

r 0 ∈π −1 (r)

n(s 0 , r 0 ). (2)

Note that equation (2) implies that π is a morphism of weighted-relation posets, and taking s = ˆ0, we see that π is also surjective.

To prove that a given rank-preserving function is a covering map, it suffices to prove equation (2) for |r| − |s| = 1 For suppose (2) holds when |r| − |s| = 1, and suppose

inductively it holds for |r| − |s| < n, n > 1 Let r, s ∈ P with |r| − |s| = n, and let π(s 0 ) = s Then

n(s, r) = X

|t|=|s|+1

n(s, t)n(t, r) = X

|t|=|s|+1

X

t 0 ∈π −1 (t)

X

r 0 ∈π −1 (r)

n(s 0 , t 0 )n(t 0 , r 0 ),

and since the sets π −1 (t), as t runs through the rank-(|s| + 1) elements of P , partition the

rank-(|s| + 1) elements of P 0,

n(s, r) = X

|t 0 |=|s 0 |+1

X

r 0 ∈π −1 (r)

n(s 0 , t 0 )n(t 0 , r 0) = X

r 0 ∈π −1 (r)

n(s 0 , r 0 ).

If P is a fixed weighted-relation poset, there is a category W/P of covers of P whose

objects are covering maps π : P 0 → P A morphism from π1 : P1 → P to π2 : P2 → P in

W/P is a morphism f : P1 → P2 in W such that π2f = π1 In fact, all such functions f

are covering maps

P2 is a morphism of weighted-relation posets such that π2f = π1 Then f is a covering map.

Proof We show f satisfies equation (2) above Let s, r ∈ P2, s 0 ∈ P1 with f (s 0 ) = s Since π2 is a covering map,

n(π2(s), π2(r)) =

k

X

i=1

n(s, r i ),

where π −12 (π2(r)) = {r1, , r k } For each r i in the image of f ,

X

r 0 ∈f −1 (r i)

n(s 0 , r 0)≤ n(s, r i ). (3)

Now Sk

i=1 f −1 (r i ) = π1−1 (π2(r)), and since π1 is a covering map we have

X

r 0 ∈π −11 (π2(r))

n(s 0 , r 0 ) = n(π2(s), π2(r)) =

k

X

i=1

n(s, r i ). (4)

Comparing (3) and (4), we see there is a contradiction unless each of the sets f −1 (r i) is

nonempty and (3) is an equality for all i.

of weighted-relation posets, with Q simple Then f can be lifted to P 0 , i.e., there is a morphism of weighted-relation posets f 0 : Q → P 0 such that πf 0 = f

Proof We define f 0 : Q → P 0 by induction on rank; there is no problem getting started

since f 0 must take ˆ0 ∈ Q to ˆ0 ∈ P 0 Suppose f 0 has already been defined for rank < n For a rank-(n − 1) element z ∈ Q and a rank-n element x ∈ f (Q) with x  f (z), let

C(x, z) = {z 0 ∈ Q| z 0  z, |z 0 | = n, and f(z 0 ) = x}.

Since the Hasse diagram of Q is a tree, sets of the form C(x, z) partition the rank-n elements of Q We shall extend f 0 to C(x, z) For z 0 ∈ C(x, z),

n(f (z), x) ≥ X

z 0 ∈C(x,z)

n(z, z 0 ) = card C(x, z).

Let S = {y ∈ P 0 | y  f 0 (z) and π(y) = x} For any y ∈ S,

n(f (z), x) = n(πf 0 (z), π(y)) = X

y 0 ∈S

n(f 0 (z), y 0)

and hence

card C(x, z) ≤

k

X

i=1

n(f 0 (z), y i ), (5)

where S = {y1, y2, , y k } Choose a partition of C(x, z) into disjoint subsets S1, , S k

(some possibly empty) so that S i has cardinality at most n(f 0 (z), y i): this is possible

because of inequality (5) Extend f 0 to C(x, z) by setting f 0 (z 0 ) = y i for all z 0 ∈ S i Then

for all z 0 ∈ C(x, z),

n(f 0 (z), f 0 (z 0))≥ X

f 0 (z 00 )=f 0 (z 0)

n(z, z 00 ).

Reasoning in the same way as in the paragraph following equation (2) above, we can

con-clude that f 0 is extended as a morphism of weighted-relation posets; and by construction

πf 0 (z 0 ) = x = f (z 0)

for all z 0 ∈ C(x, z).

π : e P → P so that e P is a simple weighted-relation poset Further, the fiber π −1 (x) of each

x ∈ P contains n(ˆ0, x) elements.

Proof Again we proceed by induction on the rank Let P (n) be the set of elements of P

of rank at most n Suppose a covering π : e P (n−1) → P (n−1)with eP (n−1)simple has already

been constructed, and let x be a rank-n element of P Since P is locally finite, the set C(x)

of elements covered by x is finite: let C(x) = {x1, , x r } Each fiber π −1 (x i) contains

n(ˆ0, x i ) rank-(n − 1) elements of e P : call them e x i1 , e x i2 , , e x im i , where m i = n(ˆ0, x i) Let

K(x) be the set

{(i, j, k)| 1 ≤ i ≤ card C(x), 1 ≤ j ≤ n(ˆ0, x i ), 1 ≤ k ≤ n(x i , x)},

and define

e

P (n)= eP (n−1) ∪ a

x∈P,|x|=n

K(x).

Extend the weight system (and order) of eP (n−1) to eP (n) by putting

n(z, (i, j, k)) =

(

1, if z  e x ij ,

0, otherwise, for any (i, j, k) ∈ K(x), z ∈ e P (n−1) Then eP (n) is simple: for any (i, j, k) ∈ K(x) there is

a unique chain to ˆ0 passing through ex ij, so

n(ˆ0, (i, j, k)) = n(ˆ0, e x ij )n(e x ij , (i, j, k)) = 1.

(The set C(x) ∩ C(x 0 ) may be nonempty for x 6= x 0, so the same point of eP (n−1) may be labelled as both ex ij and ex 0

pq, but this does not affect the conclusion since we are taking a

disjoint union of the K(x).)

Now extend π to e P (n) by having π send each element of K(x) to x Then π −1 (x) =

K(x) contains

r

X

i=1

n(ˆ0, x i )n(x i , x) = n(ˆ0, x)

elements Also, for any z ∈ e P (n−1) and rank-n element x of P , we have

n(π(z), x) =

r

X

i=1

n(π(z), x i )n(x i , x) =

r

X

i=1

m i

X

j=1

n(x,xXi)

k=1

n(z, e x ij )n(e x ij , (i, j, k)) = X

w∈π −1 (x)

n(z, w),

so π is extended as a covering map.

By a universal cover of P , we mean a cover e P → P so that, for any other cover

P 0 → P , there is a morphism ofW/P from e P → P to P 0 → P

if e P is simple, and such a cover is unique up to isomorphism in W/P

Proof Suppose that p : e P → P is a cover with e P simple, and let π : P 0 → P be another

cover By Theorem 3.2, p can be lifted to a morphism p 0 : eP → P 0 of weighted-relation

posets so that πp 0 = p: but this means p : e P → P is a universal cover Thus, a simple

cover is universal

Now suppose π 0 : P 0 → P is a universal cover By Theorem 3.3 there is a simple

cover π : e P → P , and by universality there is a morphism of W/P from π 0 to π Thus there is a morphism of weighted posets f : P 0 → e P which (by Theorem 3.1) is a covering

map, hence surjective; and since eP is simple, Proposition 2.2 says f is injective and P 0 is

simple It follows that f is an isomorphism of W/P

Example 1 Let P be the poset of subsets of {1, 2, , n}, ordered by inclusion, with each

covering relation given weight 1 Then the universal cover eP can be identified with the

set of linearly ordered subsets of {1, 2, , n}, with A  B in e P if A is an initial segment

of B; and e P → P forgets the order Evidently the fiber of any rank-k element of P has k! elements, so there are a total of k! n k

rank-k elements in e P Example 2 Let M be the poset of monomials in k commuting variables x1, , x k, with

m  m 0 in M if there is a monomial m 00 such that m 0 = mm 00 The rank on M is given

by total degree, each of the x i having degree one; the least element of M is the empty

monomial 1; and the covering relations are all given weight 1 Then the universal cover

M is isomorphic to the poset of monomials in k noncommuting variables X1, , X k, with weights given by

n(w, w 0) =

(

1, if w 0 = wX i for some i,

0, otherwise,

for |w 0 | − |w| = 1 Clearly f M is simple The function π : f M → M that sends X i to x i

(so, e.g., π −1 (x21x2) ={X2

1X2, X1X2X1, X2X12}) is a covering map The cardinality of the

fiber of any monomial is given by

n(1, x i1

1 x i2

2 · · · x i k

k ) =



i1+· · · + i k

i1 i2 · · · i k



,

and the total number of rank-n elements of f M is

X

i1+···+i k =n



n

i1 · · · i k



= k n

Example 3 Let N be the set of circular necklaces made of beads of k colors: a rank-m

element of N is a necklace with m beads, and the least element is the empty necklace ∅ For a rank-(m − 1) necklace p and a rank-m necklace q, p ≺ q if q can be obtained from

p by insertion of a bead of any color, and n(p, q) is the number of ways to insert a bead

into p to get q For example, in the case k = 2,

n( b

b

b

b b

) = 2 and n( b

b

b

b ) = 1.

The universal cover eNcan be described as the poset of necklaces with labelled beads, i.e.,

the beads of a rank-m necklace are labelled 1, 2 , m, with eN → N the function that forgets the labels It is clear that eN is simple, since there is a unique chain from any labelled necklace to ∅ via the operation of removing the highest-label bead A rank-m

element of eN can be thought of as a “k-colored permutation” mod rotation, so there are

k m (m − 1)! such elements Also, the fiber of a given necklace p ∈ N with m beads has

n(∅, p) = m!/N(p) elements, where N(p) is the number of rotations that take p to itself

(necessarily a divisor of m): p is called primitive if N(p) = 1 Evidently a necklace p with

N(p) = d has a primitive “quotient necklace” of size m

d Thus, if P (m) is the number of primitive necklaces of size m, we have

X

d|m

P ( m

d)

m!

d =

X

|p|=m

n(∅, p) = k m (m − 1)!,

orP

d|m P (d)d = k m By M¨obius inversion we obtain the classical result

P (m) = 1

m

X

d|m

µ(d)k m d

Cf [7, Theorem 7.1]

Example 4 Let B be the set of balanced bracket arrangements: a rank-n element of B

is a sequence of n left brackets and n right brackets so that, reading left to right, the number of right brackets never exceeds the number of left brackets For b, b 0 ∈ B with

|b 0 | − |b| = 1, let n(b, b 0) be the number of ways to insert a balanced pair hi into b to

obtain b 0 , e.g., n(hihi, hhiihi) = 1 and n(hihi, hihihi) = 3 The least element is the empty

arrangement ∅ ThenB is a weighted-relation poset

The universal cover eB has rank-n elements that are permutations a1a2· · · a 2n of the

multiset{1, 1, 2, 2, , n, n} such that, if a i > a j and i < j, then there is some k < j, k 6= i, with a k = a i In particular, if s is a rank-n element of eB, then the two occurrences of n in

s must be adjacent We define a partial order on eBby declaring that the rank-n element

a1a2 a 2n covers the rank-(n − 1) element a1· · · a i−1 a i+2 · · · a 2n , where a i = a i+1 = n,

and define the weight of all covering relations to be 1 Then eBis evidently simple

Define π : eB→Bby sending s ∈ eBto the bracket arrangement obtained by replacing

the first occurrence of each positive integer in s by h, and the second occurrence of each

positive integer byi Let s be a rank-(n − 1) element ofB, with π(s 0 ) = s Then a rank-n element r 0  s 0 is obtained by inserting nn into s 0, corresponding to inserting hi into s.

Thus, for any r  s in B with |r| − |s| = 1,

n(s, r) = number of ways to insert hi into s to get r

= X

r 0 ∈π −1 (r)

n(s 0 , r 0 ),

so π is a covering map.

It is well known that there are C n rank-n elements of B, where

C n = 1

n + 1



2n

n



is the nth Catalan number The number of rank-n elements of eBcan be seen to be

(2n − 1)!! = (2n − 1)(2n − 3) · · · 3 · 1

as follows If s = a1a2· · · a 2n ∈ eB, there are 2n − 1 possible choices of i so that a i is the

first occurrence of n in s Once i is chosen, then a i+1 = n, so s covers the rank-(n − 1) element a1· · · a i−1 a i+2 · · · a 2n of eB, which by induction can be chosen in (2n − 3)!! ways.

The phenomenon that labelling elements of a set enumerated by Catalan numbers gives

a set enumerated by double factorials was noted in [3]

Example 5 LetF be the set of partitions of nonnegative integers, ordered by inclusion of

their Young diagrams Thus, a partition λ of n covers a partition µ of n − 1 if λ can be obtained from µ by increasing one part of µ by 1, or by adding a new part of size 1 to µ: and we assign weight 1 to every covering relation Then a rank-n element of the universal

cover eFis a Young diagram with boxes labelled 1, 2 , n so that the labels increase from

left to right and from top to bottom, i.e., a standard Young tableau The ordering on eF

is by inclusion, and eF→F is the obvious function The cardinality n(∅, λ) of the fiber of

a partition λ is given by the hook-length formula (see [12, Cor 7.21.6]) More generally, when µ ≺ λ the number n(µ, λ) counts standard Young tableaux of skew shape λ/µ (see

[12, Cor 7.16.3] for a formula) There is also an algebraic interpretation of the numbers

n(µ, λ): if we let s λ be the Schur symmetric function corresponding to the partition λ,

then

s k1s µ = X

|λ|=|µ|+k

n(µ, λ)s λ

(see [12, Sect 7.15])

Example 6 LetCbe the poset of compositions, i.e., finite sequences of integers, with rank given by the sum, and least element∅ For compositions I, J with |J| − |I| = 1, we define

n(I, J) =







1, if J is obtained from I by increasing one part;

m, if there are m ways to insert 1 into I to get J;

0, otherwise

Thus, e.g., n(121, 122) = 1, n(121, 1121) = 2, and n(121, 212) = 0 This defines a weight

system of C, so C is a weighted-relation poset

A rank-n element of the universal cover eCis a Cayley permutation of length n as defined

in [6], i.e., a length-n sequence s of positive integers such that any positive integer i < j appears in s whenever j does The partial order on eCis defined as follows If s = a1· · · a n

is a Cayley permutation, let m(s) = max{a1, , a n } Then s covers a1· · · a n−1 if the

latter is a Cayley permutation: otherwise, s covers p(a1)· · · p(a n−1 ), where p is the

order-preserving bijection from {a1, , a n−1 } to {1, 2, , m(s) − 1} For example, the order

ideal generated by 41332 is

41332 3122  312  21  1  ∅.

If we give each covering relation weight 1, then eC is evidently simple

Let π : eC→Cbe the function that sends a sequence s to the sequence i1i2· · · i k, where

i j is the number of times j occurs in s; e.g., π −1(13) = {1222, 2122, 2212, 2221} To see

that π is a covering map, consider compositions I, J with |J| = |I| + 1 Let I = i1· · · i k

and s = a1· · · a n ∈ eCwith π(s) = I Suppose first that J is obtained from I by increasing the size of one part, so J = i1· · · i r−1 (i r + 1)i r+1 · · · i k Then n(I, J) = 1 and there is only one t  s with π(t) = J, namely t = a1· · · a n r Now suppose J is obtained from

I by inserting 1, i.e., J = i1· · · i r 1i r+1 · · · i k; without loss of generality we can assume

i r 6= 1 Then J contains a string of 1’s of length n(I, J) after i r The possible elements

t  s in eC with π(t) = J are of the form t = q(a1)q(a2)· · · q(a n )(r + i), where i runs from 1 to n(I, J) and q is the order-preserving bijection from {a1, , a n } = {1, , k} to {1, , r + i − 1, r + i + 1, , k + 1} Finally, if n(I, J) = 0, we must have n(s, t) = 0 for

any t ∈ eC with π(t) = J since the previous two cases have exhausted all the possibilities for t to cover s So in any case,

n(I, J) = X

t∈π −1 (J)

n(s, t)

when |J| − |I| = 1 and π(s) = I.

The cardinality of n(∅, I) of the fiber of a composition I = i1· · · i k is evidently the multinomial coefficient 

|I|

i1 · · · i k



.

There is an algebraic interpretation of the numbers n(I, J) analogous to that of the preceding example: if M I is the monomial quasisymmetric function corresponding to the

composition I (see [7, Sect 9.4], or [12, Sect 7.19] for definitions), then

M1k M I = X

|J|=|I|+k

n(I, J)M J

In particular, the multinomial coefficients n(∅, J) appear in the expansion of M k

1.

Example 7 LetT be the poset of rooted trees ordered by inclusion, i.e., t 0  t if t 0 can be

obtained from t by adding new edges and vertices The rank function is given by

|t| = number of vertices of t − 1,

and the least element is the tree • consisting of the root vertex The weight system is

defined as follows: if |t 0 | − |t| = 1, let n(t, t 0 ) be the number of vertices of t to which a new edge and terminal vertex may be added to obtain t 0

Rank-n elements of eT are permutations of{1, 2, , n} A permutation σ = s1s2· · · s n

of {1, , n} with s i = n covers the permutation

τ = s1· · · s i−1 s i+1 · · · s n

of {1, , n − 1} (and no other) The least element is the empty permutation ∅ Then eT

is clearly simple if we give each covering relation weight 1

Now we define the covering map π : eT → T Let π(∅) = •, and given a nonempty permutation σ = s1s2· · · s n define a rooted tree with vertices labelled 0, 1, , n as follows Label the root 0, and attach the vertex labelled i to the vertex labelled j < i if j is the last element of the sequence s1s2 s k−1 that is smaller than i, where s k = i; attach i to the root if no such j exists This associates a labelled rooted tree with each σ ∈ eT, and

π(σ) is just the rooted tree obtained by forgetting the labels Thus, e.g.,

π(4231) =

To see that π is a covering map, note first that terminal vertices of π(σ) correspond either to descents of σ (i.e., terms s i with s i > s i+1), or to the final term Now a

permutation σ with |σ| = n covers τ exactly when σ is obtained by inserting n into τ ,

e.g., 2413  213 This always introduces a new descent (or new final term) into τ, and

corresponds to adding a new edge and terminal vertex to π(τ ); moreover, the n possible